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5x⁴ – 15x² + 10x – 2x³ + 6x – 4 – (6x³ + 8x² – 10x – 3x² – 4x + 5)

= 5x⁴ – 15x² – 2x³ + 16x – 4 – 6x³ – 5x² + 14x – 5

= 5x⁴ – 8x³ – 20x² + 30x - 9

12x² + 9xy + 8xy

= 12x² + 9xy + 8xy + 6y² – 14x² + 6xy + 7xy – 3y²

= - 2x² + 30xy + 3y²

(d) 3xy² + 5y – xy²

Expression with two unlike terms is called a ‘Binomial’.

The expression 3xy² + 5y – xy² is further simplified as,

= 3xy² + 5y – xy²

= (3xy² – xy²) + 5y

= 3xy² + 5y

5x² + 10x – 3x – 6 – 8x² + 6x – 20x + 15

= - 3x² – 7x + 9

(b) – 10xyz², 3xyz²

The terms having the same algebraic factors are called like terms.

i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)

= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)

= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)

= (y – 3x) (3y – 3x) + (x + y) (x – 3y)

= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)

= 3y² – 3xy – 9xy + 9x² + x² – 3xy + xy – 3y²

= 10x² – 14xy

ii) (m + n) (m² – mn + n²)

= m(m² – mn + n²) + n(m² – mn + n²)

= m³ – m²n + n²m + nm² – mn² + n³

= m³ + n³

\((\frac{1}{3}x)^2-(\frac{y\times y}{5})^2\)

= \((\frac{1}{3}x-\frac{y\times y}{5})(\frac{1}{3}x+\frac{y\times y}{5})\)

= \(\frac{1}{9}x^2-\frac{1}{25}y^4\)

Putting x = -1 and y = -2, we have

\((\frac{1}{3}(-1)-\frac{(-2)(-25)}{5})\) = \((\frac{1}{9}(-1)^2-\frac{-2\times -2\times -2\times -2}{25})\)

= \((\frac{-1}{3}-\frac{4}{5})(\frac{-1}{3}+\frac{4}{5})=(\frac{1}{9}-\frac{16}{25})\)

= \((\frac{-17}{15})(\frac{7}{15})=\frac{-119}{225}\)

= \(\frac{-119}{225}\) = \(\frac{-119}{225}\)

Therefore,

L.H.S = R.H.S

Hence, verified

4y(3y + 4) 

= 4y × 3y + 4y × 4 

= 12y² + 6y

(x³ + 4x²y – 2xy² – 8y³) × x²y²

= x⁵y² + 4x⁴y³ – 2x³y⁴ – 8x²y⁵

x² (x² – 3xy + 2xy – 3y²)

= x² (x² – xy – 6y²)

= x⁴ – x³y – 6x²y²

i) ax² + bx + c

ii) px + qy + rz

iii) x² + y² + z²

i) 7x² and 2x are unlike terms is true. Since the power of the variable x is not same in both the terms.

ii) pq² and – 4pq² are like terms is true. Since both the terms are having same variables and same exponents.

iii) xy, -12x²y and 5xy² are like terms is false. Since all the terms are not contains same exponents.

i) No. of days ¡n the month of January are fixed in every year. So, “Number of days” is a constant. 

ii) The temperature of a day changes every minute. So (temperature) it is a variable. 

iii) The length of the classroom is fixed. So it is a constant. 

iv) The height of a growing plant changes in every month. So it is a variable.

i) Group of like terms : [a², -2a²]

ii) Group of like terms: {-yz, 2zy}

iii) Group of like terms: {-2xy², 5y²x}

iv) Group of like terms: {7p, -2p, 3p} , : {8pq,-5pq}

Given (a + b + c) + (2a + 3b – c) – (4a + b – 2c) 

= a + b + c + 2a + 3b – c – 4a – b + 2c 

= (a + 2a – 4a) + (b + 3b – b) + (c – c + 2c) 

= (1 + 2 – 4)a + (1 + 3 – 1)b + (1 – 1 + 2)c 

= (- 1) a + 3b + 2c 

= – a + 3b + 2c

Correct option is  D) 2a – 1

Correct option is (D) 2a – 1

\(\frac{4a^2-1}{2a+1}=\frac{(2a)^2-1}{2a+1}=\frac{(2a+1)(2a-1)}{(2a+1)}\) = 2a – 1

True.

The terms having the same algebraic factors are called like terms.

False.

Consider the given expression, x + y + 5x

The expression contains like terms, x + 5x = 6x

Then, the given expression becomes = y + 6x is a binomial.

Algebraic expressions are used in the following situations:

 i) Area of a triangle = 1/2 × base × height = 1/2 bh 

ii) Perimeter of a rectangle = 2(length + breadth) = 2(l + b)

The sum of 4x and 7y 

= (4x) + (7y) 

= 4x + 7y ≠ 11xy 

I do not agree with Rehman’s solution.

4n + (-3n) = n 

5pq + 12pq = 17pq 

– 5x²y + (-3x²y) = – 8x²y 

2ab² + 11ab² = 13ab²

2pq + 4pq = 6pq ≠ 8p²q²

Hence she is not right.

No.

7p + 7q

(i) Given expression is 5x² + 3y + 7

Numerical terms = 7 

Algebraic terms = 5x² , 3y

(ii) Given expression is 5x² y + 3

Numerical terms = 3 

Algebraic terms = 5x² y

(iii) Given expression is 3x² y

Numerical terms = No 

Algebraic terms = 3x² y

(iv) Given expression is 5x – 7

Numerical terms = – 7 

Algebraic terms = 5x

(v) Given expression is 7x³ – 2x

Numerical terms = No 

Algebraic terms = 7x³ , – 2x

(i) Given expression is mn² + m²n

(ii) Given expression is 7m² – 5m – 3

(iii) Given expression is 11 – 5m² + n + 8 mn

Expression given is

11x = 11 x (-5)= -55   [put x = —5]

Clearly, -55 <11.

Hence, the value is grater than 11.

Let the number be n.

So, according to the statement, the expression can be written as = 2n+1.

Yes, the result is always an odd number, because when a number becomes multiplied by 2, it becomes even and addition of 1 in that even number makes it an odd number.

The length of PR = \(\overline{PQ} + \overline{QR} \)

= 3a + 2a 

= 5a units

The perimeter of the triangle = \(\overline{AB} + \overline{BC} + \overline{CA} \)

= 2x + 6x + 5x 

= 13x units

As per the condition given in the question, p × 2q × 3r.

Correct option is :  (a) ENOD

By reversing the word from left to right. So, DONE = ENOD .

Given x = – 5 and value = – 15

Value = – 15 , 

= 3 × – 5 , 

= – 3 × x (∵ x = – 5) 

∴ Expression = 3x

From the figure, 

Given PR = 3a and RQ = 2a 

PQ = PR + RQ 

= 3a + 2a 

= (3 + 2)a 

PQ = 5a 

If a = 3 cm, then PQ = 5(3) 

∴ PQ = 15 cm

i) 3x (4ax + 8by) 

= 3x × 4ax + 3x × 8by

= 12ax² + 24bxy

ii) 4a²b (a – 3b) 

= 4a²b × a – 4a²b × 3b

= 4a³b – 12a²b²

iii) (p + 3q²) pq 

= p × pq + 3q² × pq

= p²q + 3pq³

iv) (m³ + n³) 5mn² 

= m³ × 5mn² + n³ × 5mn²

= 5m⁴n² + 5mn⁵

Quotient of x and 15 = x ÷ 15

As per the condition given the question, Quotient of x and 15 multiplied by x = (x ÷ 15)x

= x²/15

Therefore, the obtained expression is monomial.

The value of -9x when x = -3

-9x = -9 (-3) = + 27

When x = -3, 

then the value of an expression 3x is -9.

∴ 3x = 3 (-3)= -9.

Given x = 2 and value = 15 

Value = 15

= 30/2 

= 1/2 x 15 x 2

= 1/2 x 15 x \(x\) (∵ x = 2)

∴ Expression = 15x/2

x + y + z is an expression which is neither monomial nor binomial.

The given expression contains 3 terms so; it is a trinomial.

We have, -a-b-c=-a-(b+c) [by taking common (-) minus sign]

So,-a-b-c is same as -a-(b+ c).

D) Not defined

Correct option is (D) Not defined

If any term is divided by zero, then it is not defined.

\(\therefore\) \((3x^2–7x+9)\div0\) is not defined.

From the question it is given that,

x is multiplied by itself = x × x = x²

the product of x and y = x × y = xy

Then, As per the condition in the question = x² + xy

Therefore, the obtained expression is binomial.

From the question it is given that,

Three times of p = 3p

Two times of q = 2q

Three times of p and two times of q are multiplied = 3p × 2q =3p2q

Then, As per the condition in the question = r – 3p3q

Therefore, the obtained expression is binomial.

x³ – 2x² + 3x + 1

The coefficient of x² in the given expression is -2.

Coefficient is the numerical factor in a term. Sometimes, any factor in a term is called the coefficient of the remaining part of the term.

(i) Given x reduced by 5 = x – 5

(ii) Given 8 more than twice of k = 2k + 8

(iii) Given Half of y = 1/2 of y = y/2

(iv) Given One fourth of product of b and c

= 1/4 of b xc

= 1/4 x bc = bc/4

(v) Given One less than three times of p 

= 3 times of p – 1 

= 3 ∙ p – 1

Given expression is 5x when x = – 2 

5x = 5(- 2) = – 10 

Chaitanya is correct. 

Here we have to multiply 5 and,- 2. 

But Reeta subtracted. 

So, Reeta is wrong.

Correct option is  B) m² + 24m

Correct option is (B) m⁴ + 24m

\(m^3(m–2)+2m^2(m+3)–6m(m–4)\)

\(=m^4-2m^3+2m^3+6m^2-6m^2+24m\)

= \(m^4+24m\)

In row = 1, 2(1) + 1 = 3 

In row = 2, 2(2) + 1 = 5 

In row = 3, 2(3) + 1 = 7 

In row = 4, 2(4) + 1 = 9 

In row = n, 2(n) + 1 = 2n + 1

Correct option is B) \(\frac{xy}{5}\)